# Spring with a mass at an angle, need original length of spring

• 2Dark4u
In summary, the person is trying to find the length of the spring not stretched, but they can't seem to figure it out.
2Dark4u

## Homework Statement

The Setup is depicted in the Attachment. Essentially there is a 5kg Mass on an incline (45). The incline measures 0.4m from the mass to the corner were it turns 90 degrees and measures 0.3m to were there is a spring (k = 20N/m. This spring attaches to the Mass. This system is in equilibrium with the spring stretched out as it is.

The question is what is the length of the Spring not stretched. As in if the mass was un attached and the spring was at rest?

## Homework Equations

Only relevant equation i can think of is F = k(λx) Basic Hookes law

## The Attempt at a Solution

So first thing I did
5* 9.8 = 49N

Then 49cos(45) to get Fx of mass = 34.65N this should equal the X component of the force of the spring.

Second the spring should measure 0.5 since it makes a triangle right triangle with 0.4 and 0.3 this also means the angle its at in relation to the x component of force is 36.87

so F of the spring should be Fx = F cos(36.87)

34.65N / cos( 36.87) = 43.31N

then F = k λx and F/k = λx

43.31/20 = 2.16m

Now if the spring stretch measures 0.5m I don't see how λx can equal 2.16m

That would mean that the spring has a negative measurement when unstretched and that doesn't make any sense.

So help me out, were did I go wrong.

#### Attachments

• Problem 2.jpg
19.6 KB · Views: 676
2Dark4u said:

## Homework Statement

The Setup is depicted in the Attachment. Essentially there is a 5kg Mass on an incline (45). The incline measures 0.4m from the mass to the corner were it turns 90 degrees and measures 0.3m to were there is a spring (k = 20N/m. This spring attaches to the Mass. This system is in equilibrium with the spring stretched out as it is.

The question is what is the length of the Spring not stretched. As in if the mass was un attached and the spring was at rest?

## Homework Equations

Only relevant equation i can think of is F = k(λx) Basic Hookes law

## The Attempt at a Solution

So first thing I did
5* 9.8 = 49N

Then 49cos(45) to get Fx of mass = 34.65N this should equal the X component of the force of the spring.

Second the spring should measure 0.5 since it makes a triangle right triangle with 0.4 and 0.3 this also means the angle its at in relation to the x component of force is 36.87

so F of the spring should be Fx = F cos(36.87)

34.65N / cos( 36.87) = 43.31N

then F = k λx and F/k = λx

43.31/20 = 2.16m

Now if the spring stretch measures 0.5m I don't see how λx can equal 2.16m

That would mean that the spring has a negative measurement when unstretched and that doesn't make any sense.

So help me out, were did I go wrong.

I wonder if k was supposed to equal 20 kN/m

correction ! sorry

Draw a Free Body Diagram for the 5 kg block.

Ive checked with everybody. And so far what people have been able to tell me is that it just looks like the question has to be wrong

I did draw the diagram, its how i saw were all the forces were going. Unless I messed it up somehow.

Last edited:

## 1. What is Hooke's Law and how does it relate to springs with a mass at an angle?

Hooke's Law states that the force exerted by a spring is directly proportional to its extension or compression. This means that the more a spring is stretched or compressed, the greater the force it exerts. In the case of a spring with a mass at an angle, Hooke's Law still applies, but the force exerted by the spring is affected by the angle at which the mass is attached.

## 2. How do you determine the original length of a spring when a mass is attached at an angle?

The original length of a spring can be determined by using trigonometry and the angle of the mass. By measuring the displacement of the mass from the equilibrium position and using the angle, the original length of the spring can be calculated using the formula: original length = displacement / sin(angle).

## 3. Can the original length of a spring change when a mass is attached at an angle?

Yes, the original length of a spring can change when a mass is attached at an angle. This is because the force exerted by the spring is affected by the angle of the mass, which can cause the spring to stretch or compress differently than it would with a mass attached at a 90 degree angle.

## 4. How does the angle of the mass affect the spring constant?

The angle of the mass attached to a spring does not affect the spring constant. The spring constant is a measure of the stiffness of the spring and is determined by the material and physical properties of the spring itself, not by the angle at which a mass is attached.

## 5. Is the original length of a spring the same as the equilibrium position?

No, the original length of a spring is not always the same as the equilibrium position. The original length refers to the length of the spring when there is no external force or mass attached to it. The equilibrium position, on the other hand, is the position at which the spring is in balance with the external force or mass attached to it.

Replies
8
Views
699
Replies
6
Views
384
Replies
16
Views
1K
Replies
16
Views
1K
Replies
3
Views
1K
Replies
19
Views
1K
Replies
4
Views
2K
Replies
21
Views
2K
Replies
6
Views
1K
Replies
3
Views
2K